求包括n个矩形的最小矩形面积模板

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typedef double typev;const double eps = 1e-8;const int N = 50005;int sign(double d){    return d < -eps ? -1 : (d > eps);}struct point{    typev x, y;    point operator-(point d){        point dd;        dd.x = this->x - d.x;        dd.y = this->y - d.y;        return dd;    }    point operator+(point d){        point dd;        dd.x = this->x + d.x;        dd.y = this->y + d.y;        return dd;    }}ps[N];//int n, cn;double dist(point d1, point d2){    return sqrt(pow(d1.x - d2.x, 2.0) + pow(d1.y - d2.y, 2.0));}double dist2(point d1, point d2){    return pow(d1.x - d2.x, 2.0) + pow(d1.y - d2.y, 2.0);}bool cmp(point d1, point d2){    return d1.y < d2.y || (d1.y == d2.y && d1.x < d2.x);}//st1-->ed1叉乘st2-->ed2的值typev xmul(point st1, point ed1, point st2, point ed2){    return (ed1.x - st1.x) * (ed2.y - st2.y) - (ed1.y - st1.y) * (ed2.x - st2.x);}typev dmul(point st1, point ed1, point st2, point ed2){    return (ed1.x - st1.x) * (ed2.x - st2.x) + (ed1.y - st1.y) * (ed2.y - st2.y);}//多边形类struct poly{    static const int N = 50005; //点数的最大值    point ps[N+5]; //逆时针存储多边形的点,[0,pn-1]存储点    int pn;  //点数    poly() { pn = 0; }    //加进一个点    void push(point tp){        ps[pn++] = tp;    }    //第k个位置    int trim(int k){        return (k+pn)%pn;    }    void clear(){ pn = 0; }};//返回含有n个点的点集ps的凸包poly graham(point* ps, int n){    sort(ps, ps + n, cmp);    poly ans;    if(n <= 2){        for(int i = 0; i < n; i++){            ans.push(ps[i]);        }        return ans;    }    ans.push(ps[0]);    ans.push(ps[1]);    point* tps = ans.ps;    int top = -1;    tps[++top] = ps[0];    tps[++top] = ps[1];    for(int i = 2; i < n; i++){        while(top > 0 && xmul(tps[top - 1], tps[top], tps[top - 1], ps[i]) <= 0) top--;        tps[++top] = ps[i];    }    int tmp = top;  //注意要赋值给tmp！    for(int i = n - 2; i >= 0; i--){        while(top > tmp && xmul(tps[top - 1], tps[top], tps[top - 1], ps[i]) <= 0) top--;        tps[++top] = ps[i];    }    ans.pn = top;    return ans;}//求点p到st->ed的垂足，列参数方程point getRoot(point p, point st, point ed){    point ans;    double u=((ed.x-st.x)*(ed.x-st.x)+(ed.y-st.y)*(ed.y-st.y));    u = ((ed.x-st.x)*(ed.x-p.x)+(ed.y-st.y)*(ed.y-p.y))/u;    ans.x = u*st.x+(1-u)*ed.x;    ans.y = u*st.y+(1-u)*ed.y;    return ans;}//next为直线(st,ed)上的点，返回next沿(st,ed)右手垂直方向延伸l之后的点point change(point st, point ed, point next, double l){    point dd;    dd.x = -(ed - st).y;    dd.y = (ed - st).x;    double len = sqrt(dd.x * dd.x + dd.y * dd.y);    dd.x /= len, dd.y /= len;    dd.x *= l, dd.y *= l;    dd = dd + next;    return dd;}//求含n个点的点集ps的最小面积矩形，并把结果放在ds(ds为一个长度是4的数组即可,ds中的点是逆时针的)中，并返回这个最小面积。double getMinAreaRect(point* ps, int n, point* ds){    int cn, i;    double ans;    point* con;    poly tpoly = graham(ps, n);    con = tpoly.ps;    cn = tpoly.pn;    if(cn <= 2){        ds[0] = con[0]; ds[1] = con[1];        ds[2] = con[1]; ds[3] = con[0];        ans=0;    }else{        int  l, r, u;        double tmp, len;        con[cn] = con[0];        ans = 1e40;        l = i = 0;        while(dmul(con[i], con[i+1], con[i], con[l])            >= dmul(con[i], con[i+1], con[i], con[(l-1+cn)%cn])){                l = (l-1+cn)%cn;        }        for(r=u=i = 0; i < cn; i++){            while(xmul(con[i], con[i+1], con[i], con[u])                <= xmul(con[i], con[i+1], con[i], con[(u+1)%cn])){                    u = (u+1)%cn;            }            while(dmul(con[i], con[i+1], con[i], con[r])                <= dmul(con[i], con[i+1], con[i], con[(r+1)%cn])){                    r = (r+1)%cn;            }            while(dmul(con[i], con[i+1], con[i], con[l])                >= dmul(con[i], con[i+1], con[i], con[(l+1)%cn])){                    l = (l+1)%cn;            }            tmp = dmul(con[i], con[i+1], con[i], con[r]) - dmul(con[i], con[i+1], con[i], con[l]);            tmp *= xmul(con[i], con[i+1], con[i], con[u]);            tmp /= dist2(con[i], con[i+1]);            len = xmul(con[i], con[i+1], con[i], con[u])/dist(con[i], con[i+1]);            if(sign(tmp - ans) < 0){                ans = tmp;                ds[0] = getRoot(con[l], con[i], con[i+1]);                ds[1] = getRoot(con[r], con[i+1], con[i]);                ds[2] = change(con[i], con[i+1], ds[1], len);                ds[3] = change(con[i], con[i+1], ds[0], len);            }        }    }    return ans+eps;}
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